Solve 5(t)=7t^2+t^2-4

$$5 ( t ) = 7 t ^ { 2 } + t ^ { 2 } - 4$$

Solve for t

$$t = \frac{3 \sqrt{17} + 5}{16} \approx 1.085582305$$

$$t=\frac{5-3\sqrt{17}}{16}\approx -0.460582305$$

Steps Using the Quadratic Formula

Combine 7t^{2} and t^{2} to get 8t^{2}.

$$5t=8t^{2}-4$$

Subtract 8t^{2} from both sides.

$$5t-8t^{2}=-4$$

Add 4 to both sides.

$$5t-8t^{2}+4=0$$

All equations of the form ax^{2}+bx+c=0 can be solved using the quadratic formula: \frac{-b±\sqrt{b^{2}-4ac}}{2a}. The quadratic formula gives two solutions, one when ± is addition and one when it is subtraction.

$$-8t^{2}+5t+4=0$$

This equation is in standard form: ax^{2}+bx+c=0. Substitute -8 for a, 5 for b, and 4 for c in the quadratic formula, \frac{-b±\sqrt{b^{2}-4ac}}{2a}.

$$t=\frac{-5±\sqrt{5^{2}-4\left(-8\right)\times 4}}{2\left(-8\right)}$$

Square 5.

$$t=\frac{-5±\sqrt{25-4\left(-8\right)\times 4}}{2\left(-8\right)}$$

Multiply -4 times -8.

$$t=\frac{-5±\sqrt{25+32\times 4}}{2\left(-8\right)}$$

Multiply 32 times 4.

$$t=\frac{-5±\sqrt{25+128}}{2\left(-8\right)}$$

Add 25 to 128.

$$t=\frac{-5±\sqrt{153}}{2\left(-8\right)}$$

Take the square root of 153.

$$t=\frac{-5±3\sqrt{17}}{2\left(-8\right)}$$

Multiply 2 times -8.

$$t=\frac{-5±3\sqrt{17}}{-16}$$

Now solve the equation t=\frac{-5±3\sqrt{17}}{-16} when ± is plus. Add -5 to 3\sqrt{17}.

$$t=\frac{3\sqrt{17}-5}{-16}$$

Divide -5+3\sqrt{17} by -16.

$$t=\frac{5-3\sqrt{17}}{16}$$

Now solve the equation t=\frac{-5±3\sqrt{17}}{-16} when ± is minus. Subtract 3\sqrt{17} from -5.

$$t=\frac{-3\sqrt{17}-5}{-16}$$

Divide -5-3\sqrt{17} by -16.

$$t=\frac{3\sqrt{17}+5}{16}$$

The equation is now solved.

$$t=\frac{5-3\sqrt{17}}{16} t=\frac{3\sqrt{17}+5}{16}$$

Steps for Completing the Square

Combine 7t^{2} and t^{2} to get 8t^{2}.

$$5t=8t^{2}-4$$

Subtract 8t^{2} from both sides.

$$5t-8t^{2}=-4$$

Quadratic equations such as this one can be solved by completing the square. In order to complete the square, the equation must first be in the form x^{2}+bx=c.

$$-8t^{2}+5t=-4$$

Divide both sides by -8.

$$\frac{-8t^{2}+5t}{-8}=-\frac{4}{-8}$$

Dividing by -8 undoes the multiplication by -8.

$$t^{2}+\frac{5}{-8}t=-\frac{4}{-8}$$

Divide 5 by -8.

$$t^{2}-\frac{5}{8}t=-\frac{4}{-8}$$

Reduce the fraction \frac{-4}{-8} to lowest terms by extracting and canceling out 4.

$$t^{2}-\frac{5}{8}t=\frac{1}{2}$$

Divide -\frac{5}{8}, the coefficient of the x term, by 2 to get -\frac{5}{16}. Then add the square of -\frac{5}{16} to both sides of the equation. This step makes the left hand side of the equation a perfect square.

$$t^{2}-\frac{5}{8}t+\left(-\frac{5}{16}\right)^{2}=\frac{1}{2}+\left(-\frac{5}{16}\right)^{2}$$

Square -\frac{5}{16} by squaring both the numerator and the denominator of the fraction.

$$t^{2}-\frac{5}{8}t+\frac{25}{256}=\frac{1}{2}+\frac{25}{256}$$

Add \frac{1}{2} to \frac{25}{256} by finding a common denominator and adding the numerators. Then reduce the fraction to lowest terms if possible.

$$t^{2}-\frac{5}{8}t+\frac{25}{256}=\frac{153}{256}$$

Factor t^{2}-\frac{5}{8}t+\frac{25}{256}. In general, when x^{2}+bx+c is a perfect square, it can always be factored as \left(x+\frac{b}{2}\right)^{2}.

$$\left(t-\frac{5}{16}\right)^{2}=\frac{153}{256}$$

Take the square root of both sides of the equation.

$$\sqrt{\left(t-\frac{5}{16}\right)^{2}}=\sqrt{\frac{153}{256}}$$

Simplify.

$$t-\frac{5}{16}=\frac{3\sqrt{17}}{16} t-\frac{5}{16}=-\frac{3\sqrt{17}}{16}$$

Add \frac{5}{16} to both sides of the equation.

$$t=\frac{3\sqrt{17}+5}{16} t=\frac{5-3\sqrt{17}}{16}$$